Hackerrank-Problems/Cipher at master · deekshaVarshney/Hackerrank-Problems · GitHub

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#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {

   int N,K,i,j,k;
    scanf("%d %d",&N,&K);
char a[N+K];
    int b[N],c[N+K];
   // printf("%d %d\n",N,K);
    for(i=1;i<=N+K;i++)
        scanf("%c",&a[i]);
      for(i=2,k=0;a[i]!='\0';i++){
         // printf("a%c ",a[i]);
    if(a[i]=='0')
          c[k++]=0;
          else
              c[k++]=1;
      }
    //for(i=0;i<N+K-1;i++)
      //  printf("%d",c[i]);
  b[0]=c[0];
    for(i=1;i<K;i++)
        {
        b[i]=(c[i]^c[i-1]);

    }
    for(i=K;i<N;i++)
        b[i]=c[i]^c[i-1]^b[i-K];       //b[i] will become 1001010
    //printf("\n");
       for(i=0;i<N;i++)
           printf("%d",b[i]);
    return 0;

}


ex.
1 0 0 1 0 1 0
  1 0 0 1 0 1 0
    1 0 0 1 0 1 0
      1 0 0 1 0 1 0
_____________________
1 1 1 0 1 0 0 1 1 0


b[0]=c[0]=1
here in the firt loop till K(XOR of c[i-1] and c[i])
b[1]=0
b[2]=0
b[3]=1

now in the second loop from K
1. XOR of c[i-1] and c[i] will give
b[4]=1
b[5]=1
b[6]=0

2. now we Xor it with b[i] i<K-1
i.e b[0]=1
    b[1]=0
    b[2]=0
so b[4] to b[6] becomes 110 xor 100 =010

so final b becomes 1001010